Raising the 1M resistor on the input.

[valvetones]

Hello!
Been awhile , I hope you all are doing well.
So ive been playing with the EF86 normal channel on my vibroverb build. I built it with the EF86 to put the channels in phase so I could jumper channels.
This was to be able to boost gain and maybe get earlier breakup as well as to have a different tonal palate.

Well, I realized that when I jumper channels , I’m putting the two 1 Meg resistors in parallel, effectively making it 500k and bleeding half my guitar signal to ground before it even hits a grid.
This problem nearly negates the purpose of jumpering the channels in the first place.
I had considered changing the 1M resistors to 2 meg so the two parallel resistors will be 1 meg when channels are jumpered.
But then when there is no jumper and I’m actually only playing one channel, then it will be 2 meg.

What effects electrically and also sound wise might I observe if I used the 2 meg resistors , when only playing through one channel?

Thank you for your insights fellas. Hats off to the moderators.

[PRR]

Just do it. The difference is minor.

[d95err]

You are not bleeding half the signal to ground. It is only a small change of input impedance.

1M or 500k will make no significant difference to the signal level (unless there is a huge series resistance infront of it).

[nandrewjackson]

Could it also be said that the guitar signal coming in "sees" the parallel input grids as lower Z than if they were being used independently?  Or do the input grids retain their individual Z despite being parallel?

[PRR]

> do the input grids retain their individual Z despite being parallel?

Read a chapter on "parallel resistances".

[nandrewjackson]

I'm familiar with parallel resistances.


1 /  ( [1/R]+[1/R]+[1/R] )


But I was wondering if each grid being part of an active component (triode) would be an exception to the rule. As if the active component (triode) may have a different equation, a different "rule of thumb" for parallel Z, as opposed to parallel passive resistors.

[PRR]

But the 1Meg is not inside the triode.

Anyway, how would your pickup know?

[pdf64]

Pretty much the only loading that the grid presents to the input is its Miller capacitance.